Question

A body weighs (i) 900N on the surface of earth how much will it weigh on the surface of

Mars whose mass is one ninth and radius is one half that of the earth take g on the surface
of earth to be 10 m/s2

Answer 1

400 Newtons or 40 kg

Step-by-step explanation:

The gravitational force in Newtons of a body of mass mkg on the surface of the earth is given by the equation

`F_e = G\frac{mM_e} {r_e^2}`

where
`F_e` is the magnitude of the gravitational force on earth

m the mass of the object and
`M_e` the mass of the earth and
`r_e` the radius of the earth

G is the universal gravitational constant which is the same throughout the universe

On Mars, the gravitational force for the same object would be:

`F_m = G\frac{mM_m} {r_m^2}`

Since it is given that

`M_m = (1)/(9)M_e`

and

`r_m = (1)/(2)r_e`

we can rewrite
`F_m` in terms of
`F_e`

`F_m = G(m(M_(e))/(9))/(\left((r_(e))/(2)\right)^(2))`

`(4)/(9) \;G(mM_(e))/(\left(r_(e)\right)^(2)) =(4)/(9) F_e\\\\= (4)/(9) \cdot 900 = 400\;N`

Since F = mg where g = 10 m/s^2

m = F/g = 400N/10m/s² = 40 kg